(2x+5)(3x-1)+(3x-4)(2-3x)=-3x2+28x+17

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Solution for (2x+5)(3x-1)+(3x-4)(2-3x)=-3x2+28x+17 equation:



(2x+5)(3x-1)+(3x-4)(2-3x)=-3x^2+28x+17
We move all terms to the left:
(2x+5)(3x-1)+(3x-4)(2-3x)-(-3x^2+28x+17)=0
We add all the numbers together, and all the variables
-(-3x^2+28x+17)+(2x+5)(3x-1)+(3x-4)(-3x+2)=0
We get rid of parentheses
3x^2-28x+(2x+5)(3x-1)+(3x-4)(-3x+2)-17=0
We multiply parentheses ..
3x^2+(+6x^2-2x+15x-5)-28x+(3x-4)(-3x+2)-17=0
We get rid of parentheses
3x^2+6x^2-2x+15x-28x+(3x-4)(-3x+2)-5-17=0
We multiply parentheses ..
3x^2+6x^2+(-9x^2+6x+12x-8)-2x+15x-28x-5-17=0
We add all the numbers together, and all the variables
9x^2+(-9x^2+6x+12x-8)-15x-22=0
We get rid of parentheses
9x^2-9x^2+6x+12x-15x-8-22=0
We add all the numbers together, and all the variables
3x-30=0
We move all terms containing x to the left, all other terms to the right
3x=30
x=30/3
x=10

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